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# Chapter 5 Systems of Linear Equations and Inequalities - PowerPoint PPT Presentation

Chapter 5 Systems of Linear Equations and Inequalities. § 5.1. Solving Systems of Linear Equations by Graphing. Systems of Equations. We know that an equation of the form Ax + By = C is a line when graphed. Two such equations is called a system of linear equations.

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Chapter 5Systems of Linear Equations and Inequalities

§5.1

Solving Systems of Linear Equations by Graphing

We know that an equation of the form Ax + By = C

is a line when graphed.

Two such equations is called a system of linear equations.

A solution of a system of linear equations is an ordered pair

that satisfies both equations in the system.

For example, the ordered pair (2,1) satisfies the system

3x + 2y = 8

4x – 3y = 5

Blitzer, Introductory Algebra, 5e – Slide #3 Section 5.1

Since two lines may intersect in exactly one point,

may not intersect at all, or may intersect in every

point; it follows that a system of linear equations

will have exactly one solution, will have no solution,

or will have infinitely many solutions.

Blitzer, Introductory Algebra, 5e – Slide #4 Section 5.1

EXAMPLE

Determine whether (3,2) is a solution of the system

SOLUTION

Because 3 is the x-coordinate and 2 is the y-coordinate of the point(3,2), we replace x with 3 and y with 2.

?

Since the result is false, (3,2) is NOT a solution for the system. Also, I need not check the other equation since the first one failed.

?

?

false

Blitzer, Introductory Algebra, 5e – Slide #5 Section 5.1

NOTE: In order for this method to be useful, you must graph the lines veryaccurately.

Blitzer, Introductory Algebra, 5e – Slide #6 Section 5.1

EXAMPLE

Solve by graphing:

SOLUTION

1) Graph the first equation. I first rewrite the equation in slope-intercept form.

m = -4 = -4/1, b = 4

Now I can graph the equation.

Blitzer, Introductory Algebra, 5e – Slide #7 Section 5.1

CONTINUED

2) Graph the second equation on the same set of axes. I first rewrite the equation in slope-intercept form.

m = 3 = 3/1, b = -3

Now I can graph the equation.

Blitzer, Introductory Algebra, 5e – Slide #8 Section 5.1

CONTINUED

3) Determine the coordinates of the intersection point. This ordered pair is the system’s solution. Using the graph below, it appears that the solution is the point (1,0). We won’t know for sure until after we check this potential solution in the next step.

Blitzer, Introductory Algebra, 5e – Slide #9 Section 5.1

CONTINUED

4) Check the solution in both equations.

?

?

?

?

true

true

Because both equations are satisfied, (1,0) is the solution and {(1,0)} is the solution set.

Blitzer, Introductory Algebra, 5e – Slide #10 Section 5.1

Solve the system by graphing.

EXAMPLE

3x + 2y = 12

2x – y = 1

Graph the first line. Find the x and y intercepts of 3x + 2y = 12.

x-intercept y-intercept

Let y = 0 Let x = 0

3x + 2(0) =12 3(0) + 2y = 12

3x = 12 2y = 12

x = 4 y = 6

The x-intercept is 4 and the y-intercept is 6.

Blitzer, Introductory Algebra, 5e – Slide #11 Section 5.1

Graph the first line.

CONTINUED

Blitzer, Introductory Algebra, 5e – Slide #12 Section 5.1

Now, graph the second line.

CONTINUED

Find the x and y intercepts of 2x - y = 1

x-intercept y-intercept

Let y = 0 Let x = 0

2x - (0) =1 2(0) - y = 1

2x = 1 -y = 1

x = 0.5 y = -1

The x-intercept is 0.5 and the y-intercept is -1.

Blitzer, Introductory Algebra, 5e – Slide #13 Section 5.1

Graph the second line on same set of axes.

CONTINUED

Blitzer, Introductory Algebra, 5e – Slide #14 Section 5.1

Find coordinates of intersection.

CONTINUED

Point of intersection: (2,3)

Blitzer, Introductory Algebra, 5e – Slide #15 Section 5.1

3x + 2y =12

3(2)+2(3) ? 12

6 + 6 ? 12

12 = 12

True

Equation 2

2x - y =1

2(2) - (3) ? 1

4 - 3 ? 1

1 = 1

True

Systems of Equations

Check the proposed solution in each equation. It appears graphically that the solution is (2,3), but you can’t be sure until you check the point in each of the original equations.

CONTINUED

Since the point (2,3) checks in each of the original equations, it is indeed the solution of the given system of equations.

Blitzer, Introductory Algebra, 5e – Slide #16 Section 5.1

Solve the system by graphing.

EXAMPLE

y = 2x – 3

y = 2x + 7

We note that the lines have the same slope, but are not the same line. We know that they are not the same because although the slopes are the same, the y-intercepts are different. Let’s look at the graph of each of the two lines on the same set of axes.

Blitzer, Introductory Algebra, 5e – Slide #17 Section 5.1

CONTINUED

Consider the first line. Since the line, y = 2x – 3, is in slope-intercept form, we know the slope and y-intercept.

The y-intercept is -3, so the line passes through (0, -3). The slope is 2/1. To graph, we start at the y-intercept and move 2 up (the rise) and 1 right (the run).

Now, consider the second line. Since the line, y = 2x + 7, is in slope-intercept form, we know the slope and y-intercept.

The y-intercept is 7, so the line passes through (0, 7). The slope is 2/1. To graph, we start at the y-intercept and move 2 up (the rise) and 1 right (the run).

Blitzer, Introductory Algebra, 5e – Slide #18 Section 5.1

Graph the lines on the same set of axes.

CONTINUED

This system is said to be inconsistent. Since the lines are parallel and fail to intersect, the system has no solution.

We would have to say that the solution is the empty set.

Blitzer, Introductory Algebra, 5e – Slide #19 Section 5.1

Blitzer, Introductory Algebra, 5e – Slide #20 Section 5.1